✔ 1800 = 42-4264, from 51.9615. Take 30- Subt. 90030, from 42.4264. Rem. 12-4264 = 3d. share. 13. If a cubick foot of iron were hammered, or drawn, into a square bar, an inch about, that is, of an inch square: required its length, supposing there is no waste of metal ? 12x12x12 •25x-25x4 =6912 inches, = 576 feet, Ans. 14. Required the axis of a globe, whose solidity may be just equal to the area of its surface? •7854x4 ⚫5236 6 inches, Ans. 15. A joist is 7 inches wide, and 24 thick; but I want one just twice as large, which shall be 3 inches thick: what will be the breadth? 7.5x2.25x2 3.75 9 inches, Ans. 16. I have a square stick of timber 18 inches by 14; but one of a third part of the timber in it, provided it be 8 inches deep, will serve how wide will it be? 18x14 3 ÷ 8= 10 inches, Ans. 17. A had a beam of oak timber, 18 inches square throughout, and 25 feet long, which he bartered with B, for an equilateral triangular beam of the same length, each side 24 inches: required the balance at 1s. 4d. per foot? 18x18x25 144 == =56.95, solidity of the square beam. The perpendicular let fall on one of the sides of the triangular beam is 20.7846 inches, and the half perp. 10.3923, then 10.3923x24 144 1.732 foot, area at the end, and 1·732×25 = 43·3 feet, solidity of the triangular beam; therefore 56.25—43·3 = 12·95 feet, at 1s. 4d. per foot 17s. 3.2d. balance due to A, Ans. 18. What is the difference between a solid half foot, and half a foot solid? 12x 12x12x6 4, therefore, one is but of the other. 6×6×6 19. A lent B a solid stack of hay, measuring 20 feet every way; sometime afterward, B returned a quantity measuring every way 10 feet what proportion of the hay remains due? 20x20x20-10x10x10=7000 feet = Ans. 20. A ship's hold is 75 feet long, 18 wide, and 7 deep: how many bales of goods 3 feet long, 24 deep, and 24 wide, may be stowed therein, leaving a gang way the whole length, of 34 feet wide? 75-5x18 5x7-25-75.5x7-25x3.25 3.5×2.25x2.75 = 385-44 bales, Ans. 21. If a stick of timber be 20 feet long, 16 inches broad, and 8 inches thick, and 31 solid feet be sawed off one end: how long will 1728×3.5 the stick then be? 201 ---- = 16 feet, 6 inches, Ans. 16x8 22. The solid content of a square stone is found to be 136 feet; its length is 9 feet: what is the area of one end? and if the breadth be 3 feet 11 inches, what is the depth? 136.5×1728 = area 2069-0526 inches, and 2069-0.526 = 44.022 47 [ins. Ans. 9.5×12 23. I would have a cubick box made capable of receiving just 50 bushels, the bushel containing 2150-425 solid inches: what will be the length of the side? 3 2150-4x50 47.55 inches. 24. A statute bushel is to be made 8 inches high, and 18 inches diameter, to contain 2176 cubick inches; (though the content of the dimensions is but 2150-425 inches) I demand what the diameter of the bushel must be, the height being 8 inches; and what the height, the diameter being 18 inches, to contain 2176 cubick inches? Solidity. Height = 8)2176 and ✔✅ 272×1•273 = 18.6 diameter. 18.5x18.5X •7854 = 263-80315 area, and the solidity 2176÷268.8 = 8·0956 inches, height. 25. There is a garden rolling stone 66 inches in circumference, and 3 cubick feet are to be cut off from one end, perpendicular to the axis: where must the section be made? 1728×3.5 Area 412.5 14-65 inches from one end, Ans. 26. I would have a syringe of 11⁄2 inch diameter in the bore, to hold a quart, wine measure: what must be the length of the piston, sufficient to make an injection with? 1.5x1.5x-7854 = 1·76715, and 231457.75 the cubick inches 57.75 in a quart, then, 1.75715 32.679 inches, Ans. 27. If a round pillar, 9 inches diameter, contain 5 feet of what diameter is that column, of equal length, which measures 10 times as much? As &. As 5: 9x9 :: 5×10: 810, and 810 28.46 inches, Ans. 28. There is a square pyramid, each side of whose base is 30 inches, and whose perperdicular height is 120 inches, to be divided by sections parallel to its base into 3 equal parts: required the perpendicular height of each part? 30×30×40= 36000 the solidity in inches, now thereof is 24000, and is 12000. 3 As 36000 Therefore, 24000): 11520007 120x120x120 :: {12000 576000} Then, 3 : 1152000=104-8 Also, 576000=83.2. Then, 120-104-8 15.2 length of the thickest part, and 104-8-83.2 21.6 length of the middle part, consequently $3.2 is the length of the top part. 29. Suppose the diameter of the base of a conical ingot of gold to be 3 inches, and its height 9 inches; what length of wire may be expected from it, without loss of metal, the diameter of the wire being one hundredth part of an inch? 3x3x-7854×3 = 21·2058 the solidity of the cone. 21.2058 — — 270000 inch. = 4 miles, and 460 yards, Ans. •01x01×·7854 30. Suppose a pole to stand on an horizontal plane 75 feet in height above the surface: at what height from the ground must it be cut off, so as that the top of it may fall on a point 55 feet from the bottom of the pole, the end, where it was cut off, resting on the stump, or upright part? As the whole length of the pole is equal to the sum of the hypothenuse and perpendicular of a triangle, (the 55 feet on the ground being the base) this, as well as the following question, may be solved by this RULE. From the square of the length of the pole (that is, of the sum of the hypothenuse and perpendicular) take the square of the base divide the remainder by twice the length of the pole, and the quotient will be the perpendicular, or height at which it must be cut off. 31. Suppose a ship sails from latitude 43°, north, between north and east, till her departure from the meridian be 45 leauges, and the sum of her distance and difference of latitude to be 135 leagues: I demand her distance sailed, and latitude come to ? 135x135-45x45 135x2 = 60 leagues, and 60×3 = 180 miles = 3 degrees the difference of latitude, 135—60=75 leagues the distance. Now, as the vessel is sailing from the equator, and consequently the lati tude is increasing: Therefore, To the latitude sailed from Add the difference of latitude 43°,00'N. And the sum is the latitude come to = 46,00 N. AN INTRODUCTION TO ALGEBRA, DESIGNED FOR THE USE OF ACADEMIES. DEFINITIONS. ALGEBRA is the art of computing by symbols. 1. Like quantities are those which consist of the same letters: 2. Unlike quantities are those which consist of different letters: 3. Given quantities are those whose values are known. 4. Unknown quantities are those whose values are unknown. 5. Simple quantities are those which consist of one term only. 6. Compound quantities are those which consist of several terms. 7. Positive or affirmative quantities are those to be added. 8. Negative quantities are those to be subtracted. 9. Like signs are all + or all 10. Unlike signs are + and 11. The coefficient of any quantity is the number prefixed to it. 12. A binomial quantity is one consisting of two terms; a trinomial, of three terms; and a quadrinomial, of four terms, &c. 13. A residual quantity is a binomial, where one of the terms is a negative. In the computation of problems, the first letters of the alphabet are put for known quantities, and letters of the latter part of the alphabet for those which are unknown. AXIOMS. 1. If equal quantities be added to, subtracted from, multiplied or divided by, equal quantities, the wholes, remainders, products and quotients will be respectively equal. 2. The equal powers or roots of equal quantities are equal. 3. Two quantities, respectively equal to a third, are equal to each ⚫ther. 4. The whole is equal to all its parts taken together. ADDITION. CASE I. To add quantities which are alike, and have like signs.* RULE. Add all the coefficients together, and to their sum adjoin the letters common to each term, prefixing the common sign. 5a • When a leading quantity has no fign before it, 4 is always understood ; and a quantity without any coefficient prefixed to it, is fapposed to have unity, or 1. CASE II. To add quantities which are alike, but have unlike signs. RULE 1. Add all the affirmative coefficients into one sum, and all the negative ones into another. 2. Subtract the least sum from the greatest, and to the difference prefix the sign of the greatest, with the common quantity.. CASE III. To add quantities which are unlike, and have unlike signs. RULE. Collect the like quantities together by the last rule, and set down those which are unlike, one after another, with their proper signs. 2r 2r-2 RULE. Change the signs of all the quantities to be subtracted, and then add them together, as in Addition. Ba2-2b 6r2. Su+ 2 35ru-2+8r-u2 1 2a2-3b g2+9-20 24ru-8-8r-3u 1 Sar-2/ru-10 a2 + b 5r2-17a+22 11ru+6+16r-u+3u 9ar+4/ru—10—10r MULTIPLICATION. CASE I. To multiply simple quantities. RULE.-Multiply the coefficients of the two terms together, and to the product annex all the letters in those termss. Note.. |